Theorem sbbid

Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-1993) Remove dependency on ax-10 and ax-13 . (Revised by Wolf Lammen, 24-Nov-2022) Revise df-sb . (Revised by Steven Nguyen, 11-Jul-2023)

	Ref	Expression
Hypotheses	sbbid.1	|- F/ x ph
	sbbid.2	|- ( ph -> ( ps <-> ch ) )
Assertion	sbbid	|- ( ph -> ( [ y / x ] ps <-> [ y / x ] ch ) )

Proof

Step	Hyp	Ref	Expression
1		sbbid.1	|- F/ x ph
2		sbbid.2	|- ( ph -> ( ps <-> ch ) )
3	1 2	alrimi	|- ( ph -> A. x ( ps <-> ch ) )
4		spsbbi	|- ( A. x ( ps <-> ch ) -> ( [ y / x ] ps <-> [ y / x ] ch ) )
5	3 4	syl	|- ( ph -> ( [ y / x ] ps <-> [ y / x ] ch ) )